I Solving the Boat Crossing a River Problem using the Calculus of Variations

[erobz]

Now can we talk about how this works?

Choose an optimum set of paths ( for this problem ), and deviate from it by some amount $\delta$.

For this problem we demanded the total deviation from optimum be zero.

Firstly, how is that possible? I would think we should not be able to deviate at all from an optimum path with no net change in the optimized parameter.

Secondly, what happens if there is no optimum path. i.e. we try to optimize something that is not optimizable with this technique? It seems like we should have to have prior knowledge about the existence of an optimum before this can be applied?

 

[erobz]

Works for me. But one might have to hit refresh on the browser now and then

Thats right, if there is latex in the thread this will work. I forgot.

 

[kuruman]

These are good questions.

Now can we talk about how this works?

Choose an optimum set of paths ( for this problem ), and deviate from it by some amount $\delta$.

That's not what I wrote in item 1, post #12. There is only one optimum path and you start by assuming that you have found it. You give a placeholder name to a parameter that specifies it and the task is to find the value of this placeholder parameter in terms of the given quantities.

Firstly, how is that possible? I would think we should not be able to deviate at all from an optimum path with no net change in the optimized parameter.

You can calculate times for any path for a choice of $\delta$. If two different values of $\delta$ give the same time, then neither of the two paths can be optimum. However, the optimum should be between the two values of $\delta$, one of which must be positive and the other negative. Do you see why?

Secondly, what happens if there is no optimum path. i.e. we try to optimize something that is not optimizable with this technique? It seems like we should have to have prior knowledge about the existence of an optimum before this can be applied?

You cannot find something that doesn't exist. That is why before starting out to optimize a function, you need to ascertain that it can be optimized. That is what I did in post #22 when I argued that the optimum path must be at a crossing point situated between points C and D.

 

[erobz]

You can calculate times for any path for a choice of $\delta$. If two different values of $\delta$ give the same time, then neither of the two paths can be optimum. However, the optimum should be between the two values of $\delta$, one of which must be positive and the other negative. Do you see why?

Well, I think because we've chosen the coordinate $x$ to be at an extremum, that is either concave up ( local minimum ) or concave down ( local maximum ) in the vicinity of $x$. So I agree that the change in time could be zero with $\pm \delta$.

You emphasized that we deviate a small amount $\delta$. What is the significance of that in this context?

I see that $\delta \ll x$ is significant, because it can't become small arbitrarily small or else, we get $0 + 0 = 0$. This Is like a linearization in that way around $x$.